Chứng tỏ:
\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{62}+\frac{1}{63}+\frac{1}{64}>4\)
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A=1+(1/2 + 1/3 + 1/4)+(1/5 + 1/6 + 1/7 + 1/8)+(1/9+...+1/16)+(1/17+...+1/32)+(1/33+...+1/64)
A>1+(1/2 + 1/4 + 1/4)+(1/8+ 1/8+ 1/8+ 1/8)+(1/16+1/16+...+1/16)+(1/64+...+1/64)
A>1 + 1 + 1/2 + 1/2 + 1/2+ 1/2
A>4
\(=1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+...+\frac{1}{8}\right)+\left(\frac{1}{9}+...+\frac{1}{16}\right)+\left(\frac{1}{17}+...+\frac{1}{32}\right)+\left(\frac{1}{33}+...+\frac{1}{64}\right)\)
\(=1+\frac{1}{2}+\frac{1}{4}.2+\frac{1}{8}.4+\frac{1}{16}.8+\frac{1}{32}.16+\frac{1}{64}.32\)
\(=1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\)
\(=1+\frac{1}{2}.6\)
\(=1+3\)
\(=4\)
~~ Bố thí cái li.ke ~~
1 + 1/2 + 1/3 + ... + 1/62 + 1/63 + 1/64
= 1 + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + (1/9 + 1/10 + ... + 1/16) + (1/17 + 1/18 + ... + 1/32) + (1/33 + 1/34 + ... + 1/64)
> 1 + 1/2 + 1/4 × 2 + 1/8 × 4 + 1/16 × 8 + 1/32 × 16 + 1/64 × 32
> 1 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2
> 1 + 1/2 × 6
> 1 + 3
> 4
1 + 1/2 + 1/3 + ... + 1/62 + 1/63 + 1/64
= 1 + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + (1/9 + 1/10 + ... + 1/16) + (1/17 + 1/18 + ... + 1/32) + (1/33 + 1/34 + ... + 1/64)
> 1 + 1/2 + 1/4 × 2 + 1/8 × 4 + 1/16 × 8 + 1/32 × 16 + 1/64 × 32
> 1 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2
> 1 + 1/2 × 6
> 1 + 3
> 4
Ta có :
A= 1+ 1/2 + 1/3 +1/4 + ...+ 1/63 + 1/64
=1 + ( 1/2 + 1/3 + 1/4 ) + ( 1/5 +1/6 + ..+1/8 ) + ( 1/9 + 1/10 + ..+ 1/16 ) + ( 1/17 + 1/18 + ...+ 1/32 ) + ( 1/33 + 1/34 + ...+1/63 + 1/64 )
=> A > 1 + ( 1/2 + 1/4.2 ) + 1/8.4 + 1/16.8 + 1/32.16 + 1/64.32
A > 1 + 1/2 + 1/2 + 1/2 +1/2
=>A > 4